I understand how loop/global negative feedback reduces distortion, but how exactly does it reduce the output impedance? I've searched around here and on google without a definitive answer - just some general hand waving about the reduction of distortion, increase in bandwidth and reduction in output impedance.
if you really know how output distortion is reduced by negative feedback then you already know how output impedance is reduced - feedback isn't "smart" enough to know if it is correcting a nonlinear distortion or a linear voltage drop from load current flowing in device impedance - both result in a measured error at the summing input and the error is amplified by the excess loop gain (feedback factor)
Blackmans Theorem is a keystone in feedback theory - look for it in your textbook index to determine if the book is useful in really explaining feedback
Blackmans Theorem is a keystone in feedback theory - look for it in your textbook index to determine if the book is useful in really explaining feedback
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Very high feedback = necessity to use output coil = higher output impedance with frequency 😀
Quite funny.
Take a look at John Curl JC1. This power amp does really have LOW OUTPUT IMPEDANCE.
Quite funny.
Take a look at John Curl JC1. This power amp does really have LOW OUTPUT IMPEDANCE.
The amp Zout is not a single physical impedance, but rather the effect of a sagging output level with increasing load. If, say, with a load increase of 5 amps, the output level drops 1V, we say (with a wink to Mr. Ohm) that Zout = 1/5 ohms = 200milli ohms.
If you use neg feedback, the drop of the output level is 'counteracted' by the feedback (or if you prefer, compensated). So now you only have a level drop of say 100mV with a 5A output load increase. Now Zout = 0.1/5 = 20 milliohms.
BTW That's also an easy way to measure Zout. Measure the output level drop for a given output load increase, and divide.
Be aware that it is frequency dependent and possibly level dependent.
jd
If you use neg feedback, the drop of the output level is 'counteracted' by the feedback (or if you prefer, compensated). So now you only have a level drop of say 100mV with a 5A output load increase. Now Zout = 0.1/5 = 20 milliohms.
BTW That's also an easy way to measure Zout. Measure the output level drop for a given output load increase, and divide.
Be aware that it is frequency dependent and possibly level dependent.
jd
The initial purpose of negative feedback when invented was to stabilize the voltage gain, making the value of the output voltage being a multiplied value of the input voltage, and this most, in the most independant way as possible from the inside of the circuit or from the load at the output.
All the effects you describe are only consequences of this stabilisation of the gain, which is set by two resistors.
Negative feedback only lowers output impedance if it senses the output voltage, and tries to make this a defined multiple of the input voltage. You can arrange negative feedback to sense the output current instead, and then it raises output impedance. This is how an unbypassed emitter/source/cathode resistor works. You can even have a combination of the two extremes, and use feedback to set a particular impedance.
This may be why you could not find a simple explanation of why NFB reduces output impedance. Sometimes it doesn't!
This may be why you could not find a simple explanation of why NFB reduces output impedance. Sometimes it doesn't!
An unbypassed cathode/emitter res is negative voltage feedback (100% of it) and decreases Zout.
jd
I suspect janneman may be confusing emitter unbypassing with emitter following. It all depends on where you take your output from.
Of course there are different ways to use feedback, one of my favourite is Positive FB to get negative output impedance (Rout < 0).
But how many power amps use something different from pure voltage NFB ?
I think the problems Harold Black intended to solve were to stabilize voltage gain.
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